Chapter 2 Electrostatics: Problem 2.14

Problem 2.14 Find the electric field inside a sphere that carries a charge density proportional to the distance from the origin, $\rho =kr$, for some constant $k$. [Hint: This charge density is not uniform, and you must integrate to get the enclosed charge.]

Solution:

Given, $\rho =kr$ 

Consider a spherical Gaussian surface of radius $r_0$ such that $r_0\le R$ where $R$ is the radius of the sphere. 

 

 

 

 

 

 

 

 

 

 

 

By Gauss Law, $$\oint \overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{enc}}{{ϵ}_0}$$ Since the charge distribution is radially symmetrical, the direction of the electric field will be radially outwards. $$\therefore E\times 4\pi {r_0}^2=\frac{Q_{enc}}{{ϵ}_0}$$

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$$Q_{enc}={\int }_{vol.}\rho \times d\tau$$ $$Q_{enc}={\int }_{vol.}kr\times r^{2}sin\theta drd\theta d\varphi$$ $$Q_{enc}=k{\int }_0^{r_0}{r}^{3}dr{\int }_0^{\pi }sin\theta d\theta {\int }_0^{2\pi }d\varphi$$ $$Q_{enc}=k\times \frac{r^4}{4}\times 2\times 2\pi$$ 

$$Q_{enc}=k\pi {r_0}^4$$

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$$\therefore E\times 4\pi r_0^2=\frac{k\pi r_0^4}{{ϵ}_0}$$

$$\therefore E=\frac{1}{4{ϵ}_0}k{r_0}^2$$ 

$$\therefore E=\frac{1}{4\pi {ϵ}_0}\pi k{r_0}^2$$

 $$\therefore \overrightarrow{E}=\frac{1}{4\pi {ϵ}_0}\pi k{r_0}^2\hat{r}$$

 

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